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Homework Help: Desperately need introductory CIRCUIT/RESISTOR help

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    You’re given a 12.0V battery, 3 light bulbs (of 10.0Ω each), and LOTS of 10.0Ω resistors. Make a circuit so that the light bulbs’ brightnesses are multiples of 1, 2, 3. In other words, if the dimmest light bulb has brightness B, the second dimmest will have brightness 2B, and the brightest will be 3B.

    2. Relevant equations

    I've derived a few over a few hours and have drawn it out and mapped and graphed v^2/r (I'm assuming in this case brightness means POWER which is vsquared over r)

    3. The attempt at a solution
    I just don't see how this is possible to actually map out these 10 ohm resistors with a 10 ohm bulb and a set 12 volt battery. I am under the presumption that by brightness my teacher means power, and having graphed the voltages after crossing a list of quantities of 10 ohm resistors, I cannot find 3 powers that are direct multiples of 2 and 3 of one resistor configuration.

    Note that there is no statement for what we are defining as brightness, we have not been taught anything about brightness and thats actually the first time the word has popped up in this course. I don't understand because under my understanding, anything you can construct on a resistor circuit is not linearly related to the brightness, but rather the measurement of lumens of the bulb. Would appreciate if anoyne helped
    Last edited: Jun 1, 2013
  2. jcsd
  3. Jun 1, 2013 #2


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    hi martinlematre! :smile:

    hint: parallel :wink:
  4. Jun 1, 2013 #3
    I am using parallel circuits. The powers (V^2 over Resistors) are not linearly scaling with 10 ohm resistors
  5. Jun 1, 2013 #4


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    ?? :confused:

    V will be the same for parallel circuits
  6. Jun 1, 2013 #5
    There's a voltage drop across a resistor whose specific quantity is proportional to power (What I'm assuming is what my teacher means by brightness).

    V^2/R will be the variable Voltage after drops (And before each lightbulb) squared, over 10ohms (The lightbulb.

    Except there is no configuration where that variable voltage after going through X resistors has a 1, 2x and 3x value for power
  7. Jun 1, 2013 #6


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    the question says "LOTS" of resistors, so i think you're expected to get only approximately 1:2:3 :wink:

    (and it would be more logical to use I, not V)
  8. Jun 1, 2013 #7
    except im expected to draw/use a very limited program where i cant fit more than 5 or 6 resistors in a circuit without it crashing.

    im just making brightness proportional to I.
  9. Jun 1, 2013 #8


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    Suppose you had 3 light bulbs in series across the 12 volt battery.

    Then you rewire the circuit to have 1 light bulb in series with two 10 ohm resistors.

    Would the first circuit be exactly 3 times as bright as the second?
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